Introduction to Thermodynamics
So what kind of intuition do we have about heat and temperature
and energy?
Discuss the DCI.
Introduction to Thermodynamics
DCI15.1.
Two containers of water are at 20
˚C initially. One contains 50 mLs and the other
100 mLs. They are each heated with the same
source of heat for the same amount of time. If
the final temperature of the 50 mLs sample is 50
˚C what would be the final temperature of the
100 mLs sample?
A.
B.
C.
D.
E.
50 ˚C
80 ˚C
25 ˚C
100 ˚C
35 ˚C
Introduction to Thermodynamics
DCI15.1. Two containers of water are at 20 ˚C
initially. One contains 50 mLs and the other 100
mLs. They are each heated with the same source
of heat for the same amount of time. If the final
temperature of the 50 mLs sample is 50 ˚C what
would be the final temperature of the 100 mLs
sample?
A.
B.
C.
D.
E.
50 ˚C
80 ˚C
25 ˚C
100 ˚C
35 ˚C
Same amount of heat to both
beakers, but different mass. ∆T =
30˚ for beaker on the left, so ∆T is
half or 15˚.
Mass  1/∆T
Introduction to Thermodynamics
Two containers each have 50 mLs of water at 20
˚C initially. They are each heated with the same
source of heat. One is heated for ten minutes
and the other for five minutes. If the container
that was heated for five minutes has a final
temperature 30 ˚C what would be the final
temperature of the other sample?
A.
B.
C.
D.
E.
35 ˚C
40 ˚C
60 ˚C
25 ˚C
30 ˚C
Introduction to Thermodynamics
Two containers each have 50 mLs of water at 20
˚C initially. They are each heated with the same
source of heat. One is heated for ten minutes
and the other for five minutes. If the container
that was heated for five minutes has a final
temperature 30 ˚C what would be the final
temperature of the other sample?
A.
B.
C.
D.
E.
35 ˚C
40 ˚C
60 ˚C
25 ˚C
30 ˚C
Both beakers contain the same
amount of water. Twice the heat to
one. ∆T is 10˚ for smaller amount of
heat, than ∆T = 20˚ for larger amount.
Q(heat)  ∆T
Introduction to Thermodynamics
Two containers of water are at 20 ˚C initially.
One contains 50 g of water and is heated by a
source for a specified time to a final temperature
of 30 ˚C. The second container has an unknown
amount of water and is heated with the same
source to 30 ˚C. However, it takes twice as long
to get to this final temperature. How much water
is in this container?
A.
B.
C.
D.
E.
100 g
25 g
30 g
50 g
75 g
Introduction to Thermodynamics
Two containers of water are at 20 ˚C initially.
One contains 50 g of water and is heated by a
source for a specified time to a final temperature
of 30 ˚C. The second container has an unknown
amount of water and is heated with the same
source to 30 ˚C. However, it takes twice as long
to get to this final temperature. How much water
is in this container?
A.
B.
C.
D.
E.
100 g
25 g
30 g
50 g
75 g
Twice the heat is added to one beaker
to reach the same final temperature
(∆T). So the beaker must have twice
the mass.
Q(heat)  Mass
Introduction to Thermodynamics
So we have established the following relationships;
Mass  1/∆T
q(heat)  ∆T
q(heat)  mass
So
q(heat)  mass · ∆T
Heat is directly proportional to the mass times the
change in temperature.
Introduction to Thermodynamics
50 mLs of water at 80 ˚C is added to 50 mLs of
water at 20 ˚C. What would be the final
temperature?
A.
B.
C.
D.
E.
60 ˚C
40 ˚C
30 ˚C
20 ˚C
50 ˚C
Introduction to Thermodynamics
50 mLs of water at 80 ˚C is added to 50 mLs of
water at 20 ˚C. What would be the final
temperature?
A.
B.
C.
D.
E.
60 ˚C
40 ˚C
30 ˚C
20 ˚C
50 ˚C
Introduction to Thermodynamics
50 mLs of water at 80 ˚C is added to 50 mLs of
water at 20 ˚C. What would be the final
temperature?
A.
B.
C.
D.
E.
60 ˚C
40 ˚C
30 ˚C
20 ˚C
50 ˚C
qhot water + qcold water = 0
qhot water = –qcold water
masshot water · ∆Thot water = –masscold water · ∆Tcold water
50. g · ∆Thot water = –50. g · ∆Tcold water
50. g · (Tfinal – 80.0˚) = –50. g · (Tfinal – 20.0˚)
2Tfinal = 100˚ Tfinal = 50˚
Introduction to Thermodynamics
50 mLs of water at 80 ˚C is added to 100 mLs of
water at 20 ˚C. What would be the final
temperature?
A.
B.
C.
D.
E.
70 ˚C
40 ˚C
30 ˚C
60 ˚C
50 ˚C
Introduction to Thermodynamics
50 mLs of water at 80 ˚C is added to 100 mLs of
water at 20 ˚C. What would be the final
temperature?
A.
B.
C.
D.
E.
70 ˚C
40 ˚C
30 ˚C
60 ˚C
50 ˚C
Introduction to Thermodynamics
50 mLs of water at 80 ˚C is added to 100 mLs of
water at 20 ˚C. What would be the final
temperature?
A.
B.
C.
D.
E.
70 ˚C
40 ˚C
30 ˚C
60 ˚C
50 ˚C
qhot water = –qcold water
masshot water · ∆Thot water = –masscold water · ∆Tcold water
50. g · ∆Thot water = –100. g · ∆Tcold water
50. g · (Tfinal – 80.0˚) = –100. g · (Tfinal – 20.0˚)
(Tfinal – 80.0˚) = –2 · (Tfinal – 20.0˚)
3Tfinal = 120˚ Tfinal = 40˚
Introduction to Thermodynamics
50 g of water at 80 ˚C is added to 50 g of ethyl
alcohol at 20 ˚C. What would be the
approximate final temperature?
A.
B.
C.
D.
E.
60 ˚C
40 ˚C
30 ˚C
20 ˚C
50 ˚C
Introduction to Thermodynamics
50 g of water at 80 ˚C is added to 50 g of ethyl
alcohol at 20 ˚C. What would be the
approximate final temperature?
A.
B.
C.
D.
E.
60 ˚C
40 ˚C
30 ˚C
20 ˚C
50 ˚C
TWO DIFFERENT SUBSTANCES!
Experimentally the final temperature is determined to be close to 60˚.
Introduction to Thermodynamics
q(heat)  mass · ∆T
How do we make this an equality?
We must introduce a constant….in this case the
constant is called the specific heat, SH,
q(heat) = mass · SH · ∆T
Specific heat is the amount of heat required to raise the
temperature of 1 gram of a substance 1 ˚C.
Introduction to Thermodynamics
Specific Heats of Substances
Compound
Specific Heat
(J ˚C-1g-1)
H2O(l)
H2O(s)
Al(s)
C(s)
Fe(s)
Hg(l)
O2(g)
CH3CH2OH
4.184
2.03
0.89
0.71
0.45
0.14
0.917
2.46